Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r - 3}{r - 7} \times \dfrac{r^2 - 5r - 14}{-5r + 15} $
First factor the quadratic. $q = \dfrac{r - 3}{r - 7} \times \dfrac{(r - 7)(r + 2)}{-5r + 15} $ Then factor out any other terms. $q = \dfrac{r - 3}{r - 7} \times \dfrac{(r - 7)(r + 2)}{-5(r - 3)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r - 3) \times (r - 7)(r + 2) } { (r - 7) \times -5(r - 3) } $ $q = \dfrac{ (r - 3)(r - 7)(r + 2)}{ -5(r - 7)(r - 3)} $ Notice that $(r - 3)$ and $(r - 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(r - 3)}(r - 7)(r + 2)}{ -5\cancel{(r - 7)}(r - 3)} $ We are dividing by $r - 7$ , so $r - 7 \neq 0$ Therefore, $r \neq 7$ $q = \dfrac{ \cancel{(r - 3)}\cancel{(r - 7)}(r + 2)}{ -5\cancel{(r - 7)}\cancel{(r - 3)}} $ We are dividing by $r - 3$ , so $r - 3 \neq 0$ Therefore, $r \neq 3$ $q = \dfrac{r + 2}{-5} $ $q = \dfrac{-(r + 2)}{5} ; \space r \neq 7 ; \space r \neq 3 $